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| In [[mathematics]], a '''Cauchy–Euler equation''' (also known as the '''Euler–Cauchy equation''', or simply '''Euler's equation''') is a [[linear differential equation|linear]] [[homogeneous differential equation|homogeneous]] [[ordinary differential equation]] with [[variable coefficient]]s. It is sometimes referred to as an equidimensional equation. Because of its simple structure the equation can be replaced with an equivalent equation with [[constant coefficient]]s which can then be solved explicitly.
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| ==The equation==
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| Let ''y''<sup>(''n'')</sup>(''x'') be the ''n''th derivative of the unknown function ''y''(''x''). Then a '''Cauchy–Euler equation''' of order ''n'' has the form
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| :<math>a_{n} x^n y^{(n)}(x) + a_{n-1} x^{n-1} y^{(n-1)}(x) + \cdots + a_0 y(x) = 0.</math>
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| The substitution <math>x = e^u</math> reduces this equation to a linear differential equation with constant coefficients. Alternatively a trial solution <math>y = x^m</math> may be used to solve for the basis solutions.<ref name="kreyszig">{{cite book|last=Kreyszig|first=Erwin|title=Advanced Engineering Mathematics|publisher=Wiley|date=May 10, 2006|isbn=978-0-470-08484-7}}</ref>
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| ===Second order – solving through trial solution===
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| [[File:Euler-Cauchy equation solution curves real roots.png|thumb|right|Typical solution curves for a second-order Euler–Cauchy equation for the case of two real roots]]
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| [[File:Euler-Cauchy equation solution curves double root.png|thumb|right|Typical solution curves for a second-order Euler–Cauchy equation for the case of a double root]]
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| [[File:Euler-Cauchy equation solution curves complex roots.png|thumb|right|Typical solution curves for a second-order Euler–Cauchy equation for the case of complex roots]]
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| The most common Cauchy–Euler equation is the second-order equation, appearing in a number of physics and engineering applications, such as when solving [[Laplace's equation]] in polar coordinates. It is given by the equation:<ref name="kreyszig" />
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| :<math>x^2\frac{d^2y}{dx^2} + ax\frac{dy}{dx} + by = 0. \,</math>
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| We assume a trial solution given by<ref name="kreyszig" />
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| :<math>y = x^m. \,</math>
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| Differentiating, we have:
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| :<math>\frac{dy}{dx} = mx^{m-1} \,</math>
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| and
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| :<math>\frac{d^2y}{dx^2} = m(m-1)x^{m-2}. \,</math>
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| Substituting into the original equation, we have:
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| :<math>x^2( m(m-1)x^{m-2} ) + ax( mx^{m-1} ) + b( x^m ) = 0 \,</math>
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| Or rearranging gives:
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| :<math>m^2 + (a-1)m + b = 0. \,</math>
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| We then can solve for ''m''. There are three particular cases of interest:
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| * Case #1: Two distinct roots, ''m''<sub>1</sub> and ''m''<sub>2</sub>
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| * Case #2: One real repeated root, ''m''
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| * Case #3: Complex roots, α ± β''i''
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| In case #1, the solution is given by:
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| :<math>y = c_1 x^{m_{1}} + c_2 x^{m_2} \,</math>
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| In case #2, the solution is given by
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| :<math>y = c_1 x^m \ln(x) + c_2 x^m \,</math>
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| To get to this solution, the method of [[reduction of order]] must be applied after having found one solution ''y'' = ''x''<sup>''m''</sup>.
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| In case #3, the solution is given by:
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| :<math>y = c_1 x^\alpha \cos(\beta \ln(x)) + c_2 x^\alpha \sin(\beta \ln(x)) \,</math>
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| :<math>\alpha = \mathop{\rm Re}(m)\,</math>
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| :<math>\beta = \mathop{\rm Im}(m)\,</math>
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| For <math>c_1\,</math> and <math>c_2\,</math> in the real plane
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| This form of the solution is derived by setting ''x'' = ''e''<sup>''t''</sup> and using [[Euler's formula]]
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| ===Second order – solution through change of variables===
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| :<math>x^2\frac{d^2y}{dx^2} + ax\frac{dy}{dx} + by = 0 \,</math>
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| We operate the variable substitution defined by
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| :<math>t = \ln(x). \,</math>
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| :<math>y(x) = \phi(\ln(x)) = \phi(t). \,</math>
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| Differentiating:
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| :<math>\frac{dy}{dx}=\frac{1}{x}\frac{d\phi}{dt}</math>
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| :<math>\frac{d^2y}{dx^2}=\frac{1}{x^2}\bigg(\frac{d^2\phi}{dt^2}-\frac{d\phi}{dt}\bigg).</math>
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| Substituting <math>\phi(t)</math>, we have
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| :<math>\frac{d^2\phi}{dt^2} + (a-1)\frac{d\phi}{dt} + b\phi = 0. \,</math>
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| This equation in <math>\phi(t)</math> can be easily solved using its characteristic polynomial
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| :<math>\lambda^2 + (a-1)\lambda +b = 0.</math>
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| Now, if <math>\lambda_1</math> and <math>\lambda_2</math> are the roots of this polynomial, we analyze the two main cases: distinct roots and double roots:
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| If the roots are distinct, the general solution is given by
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| :<math>\phi(t)=c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}</math>, where the exponentials may be complex.
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| If the roots are equal, the general solution is given by
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| :<math>\phi(t)=c_1 e^{\lambda_1 t} + c_2 t e^{\lambda_1 t}.</math>
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| In both cases, the solution <math>y(x)</math> may be found by setting <math>t=\ln(x)</math>, hence <math>\phi(\ln(x)) = y(x)</math>.
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| Hence, in the first case,
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| :<math>y(x)=c_1 x^{\lambda_1} + c_2 x^{\lambda_2}</math>,
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| and in the second case,
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| :<math>y(x)=c_1 x^{\lambda_1} + c_2 \ln(x) x^{\lambda_1}.</math>
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| ===Example===
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| Given
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| : <math>x^2u''-3xu'+3u=0\,,</math>
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| we substitute the simple solution ''x''<sup>α</sup>:
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| : <math>x^2(\alpha(\alpha-1)x^{\alpha-2})-3x(\alpha x^{\alpha-1})+3x^\alpha=\alpha(\alpha-1)x^\alpha-3\alpha x^\alpha+3x^\alpha = (\alpha^2-4\alpha+3)x^\alpha = 0\,.</math>
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| For ''x''<sup>α</sup> to be a solution, either ''x'' = 0, which gives the [[Trivial (mathematics)|trivial]] solution, or the coefficient of ''x''<sup>α</sup> is zero. Solving the quadratic equation, we get ''α'' = 1, 3. The general solution is therefore
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| : <math>u=c_1 x+c_2 x^3\,.</math>
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| ==Difference equation analogue==
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| There is a [[difference equation]] analogue to the Cauchy–Euler equation. For a fixed ''m'' > 0, define the sequence ''ƒ''<sub>''m''</sub>(''n'') as
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| : <math>f_m(n) := n(n+1)\cdots (n+m-1).</math>
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| Applying the difference operator to <math>f_m</math>, we find that
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| : <math>
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| \begin{align}
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| Df_m(n) & = f_{m}(n+1) - f_m(n) \\
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| & = m(n+1)(n+2) \cdots (n+m-1) = \frac{m}{n} f_m(n).
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| \end{align}
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| </math>
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| If we do this ''k'' times, we will find that
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| : <math>
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| \begin{align}
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| f_m^{(k)}(n) & = \frac{m(m-1)\cdots(m-k+1)}{n(n+1)\cdots(n+k-1)} f_m(n) \\
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| & = m(m-1)\cdots(m-k+1) \frac{f_m(n)}{f_k(n)},
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| \end{align}
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| </math>
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| where the superscript <sup>(''k'')</sup> denotes applying the difference operator ''k'' times. Comparing this to the fact that the ''k''-th derivative of ''x''<sup>''m''</sup> equals
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| : <math>m(m-1)\cdots(m-k+1)\frac{x^m}{x^k}</math>
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| suggests that we can solve the ''N''-th order difference equation
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| : <math>f_N(n) y^{(N)}(n) + a_{N-1} f_{N-1}(n) y^{(N-1)}(n) + \cdots + a_0 y(n) = 0,</math>
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| in a similar manner to the differential equation case. Indeed, substituting the trial solution
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| : <math>y(n) = f_m(n) \, </math>
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| brings us to the same situation as the differential equation case,
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| : <math>m(m-1)\cdots(m-N+1) + a_{N-1} m(m-1) \cdots (m-N+2) + \cdots + a_1 m + a_0 = 0.</math>
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| One may now proceed as in the differential equation case, since the general solution of an ''N''-th order linear difference equation is also the linear combination of ''N'' linearly independent solutions. Applying reduction of order in case of a multiple root ''m''<sub>1</sub> will yield expressions involving a discrete version of ln,
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| : <math>\varphi(n) = \sum_{k=1}^n \frac{1}{k - m_1}.</math>
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| (Compare with: <math>\ln (x - m_1) = \int_{1+m_1}^x \frac{1}{t - m_1} \, dt.</math>)
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| In cases where fractions become involved, one may use
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| : <math>f_m(n) := \frac{\Gamma(n+m)}{\Gamma(n)}</math>
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| instead (or simply use it in all cases), which coincides with the definition before for integer ''m''.
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| ==See also==
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| * [[Hypergeometric differential equation]]
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| * [[Cauchy–Euler operator]]
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| ==References==
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| <references />
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| ==Bibliography==
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| *{{Mathworld|EulerDifferentialEquation}}
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| {{DEFAULTSORT:Cauchy-Euler Equation}}
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| [[Category:Ordinary differential equations]]
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